Input signal to a common emitter amplifier having a voltage gain of $1000$ is given by $v_i = (0.004 \, V) \sin(\omega t + \pi/2)$. The corresponding output signal is

In case of a bipolar transistor, $\beta = 45$. The potential drop across the collector resistance of $1 \ k\Omega$ is $5 \ V$. The base current is approximately: (in $\mu A$)

$10^{10}$ electrons enter the emitter of a junction transistor in a time of $0.4 \mu s$. If $5 \%$ of the electrons are lost in the base, then the collector current is (in $\text{ mA}$)

An $NPN$ transistor is used in common emitter configuration as an amplifier with $1\,k\Omega$ load resistance. $A$ signal voltage of $10\,mV$ is applied across the base-emitter. This produces a $3\,mA$ change in the collector current and $15\,\mu A$ change in the base current of the amplifier. The input resistance and voltage gain are:

In the common-base configuration,a transistor has a current amplification factor of $0.95$. If the transistor is used in a common-emitter configuration and the base current changes by $2 \mu A$,then the change in the collector current is: (in $\mu A$)

$A$ common emitter transistor amplifier has a current gain of $50$. If the load resistance is $9 \, k\Omega$ and the input resistance is $500 \, \Omega$,the voltage gain of the amplifier will be: